## mathmath333 one year ago find the range of x.

1. mathmath333

\large \color{black}{\begin{align}& \{a,b\}\in \mathbb{R^{>0}},\ \ a>b,\ \ \normalsize \text{then }\ \ a^{1/x}>b^{1/x} \hspace{.33em}\\~\\ & \normalsize \text{find the range of }\ x .\hspace{.33em}\\~\\ \end{align}}

2. mathmath333

i think x>0

3. mathmath333
4. ganeshie8

$a^{1/x}\gt b^{1/x}$ divide $$b^{1/x}$$ both sides and get $\left(\frac{a}{b}\right)^{1/x}\gt 1$ taking log both sides $x*\ln\left(\frac{a}{b}\right) \gt 0$ since $$\ln\left(\frac{a}{b}\right) \gt 0$$ for $$a\gt b$$, dividing it both sides wont flip the signs : $x\gt 0$

5. mathmath333

but after taking $$\ln$$ on both sides it should be this ? $$\dfrac1x \times \ln\left(\dfrac{a}{b}\right) \gt 0$$

6. ganeshie8

Ahh sry, it was just a typo

7. ganeshie8

fixed : $a^{1/x}\gt b^{1/x}$ divide $$b^{1/x}$$ both sides and get $\left(\frac{a}{b}\right)^{1/x}\gt 1$ taking log both sides $\frac{1}{x}*\ln\left(\frac{a}{b}\right) \gt 0$ since $$\ln\left(\frac{a}{b}\right) \gt 0$$ for $$a\gt b$$, dividing it both sides wont flip the signs : $\frac{1}{x}\gt 0$ Multipilying $$x^2$$ both sides gives $x\gt 0$